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Suppose that a die is rolled twice. What are the possible values that the following random variables can take on (a) the maximum value to appear in the two rolls; (b) the minimum value to appear on the two rolls; (c) the sum of the two rolls; (d) the value of the first roll minus the value of the second roll?

Short Answer

Expert verified
(a) Max: 1, 2, 3, 4, 5, 6; (b) Min: 1, 2, 3, 4, 5, 6; (c) Sum: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; (d) Difference: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

Step by step solution

01

Create a table with the possible outcomes of the two rolls

To better visualize the possible outcomes of two rolls of a die, we can create a 6x6 table. In this table, rows represent the value of the first roll, and the columns represent the value of the second roll. Let's call the value of the first roll R1, and the value of the second roll R2. R1 \ R2 | 1 | 2 | 3 | 4 | 5 | 6 -------|----|----|-----|-----|-----|----- 1 | (1,1)| (1,2)| (1,3)| (1,4)| (1,5)| (1,6) 2 | (2,1)| (2,2)| (2,3)| (2,4)| (2,5)| (2,6) 3 | (3,1)| (3,2)| (3,3)| (3,4)| (3,5)| (3,6) 4 | (4,1)| (4,2)| (4,3)| (4,4)| (4,5)| (4,6) 5 | (5,1)| (5,2)| (5,3)| (5,4)| (5,5)| (5,6) 6 | (6,1)| (6,2)| (6,3)| (6,4)| (6,5)| (6,6) Step 2: Finding the possible values for each random variable
02

(a) Maximum value of the two rolls

In this case, we are finding the maximum number rolled between R1 and R2. For any given cell (i, j) in the table, which corresponds to a roll of i on R1 and a roll of j on R2, the maximum value can be found by taking the maximum between i and j. We can compute the maximum value for each cell in the table and obtain the following table: max(R1, R2) | ------------| 1 | 2 | 3 | 4 | 5 | 6 2 | 2 | 3 | 4 | 5 | 6 3 | 3 | 3 | 4 | 5 | 6 4 | 4 | 4 | 4 | 5 | 6 5 | 5 | 5 | 5 | 5 | 6 6 | 6 | 6 | 6 | 6 | 6 From this table, we see that the possible maximum values are 1, 2, 3, 4, 5, and 6.
03

(b) Minimum value of the two rolls

Now, let's find the minimum value of the two rolls. For each cell (i, j) in the table, we take the minimum between R1 = i and R2 = j: min(R1, R2) | ------------| 1 | 1 | 1 | 1 | 1 | 1 1 | 2 | 2 | 2 | 2 | 2 1 | 2 | 3 | 3 | 3 | 3 1 | 2 | 3 | 4 | 4 | 4 1 | 2 | 3 | 4 | 5 | 5 1 | 2 | 3 | 4 | 5 | 6 The possible minimum values are 1, 2, 3, 4, 5, and 6.
04

(c) Sum of the two rolls

To find the sum of the two rolls, we simply add R1 and R2 for each cell in the original table: R1 + R2 | -----------| 2 | 3 | 4 | 5 | 6 | 7 3 | 4 | 5 | 6 | 7 | 8 4 | 5 | 6 | 7 | 8 | 9 5 | 6 | 7 | 8 | 9 | 10 6 | 7 | 8 | 9 | 10 | 11 7 | 8 | 9 | 10 | 11 | 12 The possible sums are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.
05

(d) First roll minus the second roll

Now let's find the value of R1 - R2 for each cell in the original table: R1 - R2 | ---------| 0 | -1 | -2 | -3 | -4 | -5 1 | 0 | -1 | -2 | -3 | -4 2 | 1 | 0 | -1 | -2 | -3 3 | 2 | 1 | 0 | -1 | -2 4 | 3 | 2 | 1 | 0 | -1 5 | 4 | 3 | 2 | 1 | 0 The possible values for R1 - R2 are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, and 5. In conclusion, the potential values for each random variable are: (a) Maximum value of the two rolls: 1, 2, 3, 4, 5, 6 (b) Minimum value of the two rolls: 1, 2, 3, 4, 5, 6 (c) Sum of the two rolls: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 (d) First roll minus the second roll: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
When diving into the realm of probability theory, one of the most fundamental concepts is that of random variables. A random variable, essentially, is a variable whose values result from the outcome of a random phenomenon. It's a way to quantify uncertainty, assigning numerical values to the outcomes of a random process.

Take a die roll, for instance. Each possible outcome (1 through 6) is a realization of the random variable. In our exercise where a die is rolled twice, we can analyze random variables such as the maximum and minimum values, the sum, or the difference in outcomes of the rolls. These outcomes are not known ahead of the rolls, thus they represent core examples of random variables in a probabilistic setting.

In understanding random variables, it is vital to distinguish between the two types: discrete and continuous. Discrete variables, like in our die roll example, take on a countable number of distinct values. Conversely, continuous variables can assume any value within an interval or collection of intervals. This distinction is paramount as it guides the mathematical treatment and probability distribution applications associated with the variables.
Discrete Probability Distributions
Next, we look into discrete probability distributions, which delineate how probabilities are assigned to the values of a discrete random variable. This distribution is a breakdown of the probabilities that a discrete random variable will take on each of its possible values.

An exemplary discrete distribution is the probability distribution for the outcome of a single die roll; where each of the six outcomes has an equal probability of 1/6. However, when a die is rolled twice, and a new random variable emerges, say, the sum of the die rolls, the distribution shifts. Unlike having equal chances for each outcome, certain sums become more likely than others. In fact, the sum '7' holds the crown for the highest probability due to its numerous combinations (like 1+6, 2+5, etc.).

When constructing a discrete probability distribution, we must satisfy two conditions. The probability of each value must be between 0 and 1, and the total of all probabilities for all possible values must equal 1. In our die-rolling exercise, these rules help us determine how likely specific scenarios are and are displayed through the use of probability distribution tables or formulas.
Probability Outcomes
Acknowledging the significance of probability outcomes is crucial in apprehending probability theory. Simply put, these outcomes are the potential results of a random process, that either have already occurred or may occur in a future event.

In our example, the outcomes are essentially the results of rolling a die twice. For a single roll, there are six possible outcomes—each side of the die. With two rolls, there are a total of 36 outcomes, calculated by multiplying the possible outcomes of each roll (6 from the first and 6 from the second). In probability terms, each possible couple of numbers (such as (1,1), (1,2), ... (6,6)) is a unique outcome within this sample space.

When we identify the maximum, minimum, sum, or difference of the two rolls, we are essentially creating random variables as functions of these outcomes. Understanding these outcomes and appreciating their role in the bigger picture of probability theory is a valuable step in mastering the subject. By knowing the potential results and their associated probabilities, one can calculate the likelihood of different events and thereby gauge expectations accordingly.

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Most popular questions from this chapter

A jar contains \(m+n\) chips, numbered \(1,2, \ldots, n+m\). A set of size \(n\) is drawn. If we let \(X\) denote the number of chips drawn having numbers that exceed all the numbers of those remaining, compute the probability mass function of \(X\).

Suppose that \(n\) independent tosses of a coin having probability \(p\) of coming up heads are made. Show that the probability that an even number of heads results is \(\frac{1}{2}\left[1+(q-p)^{n}\right]\), where \(q=1-p\). Do this by proving and then utilizing the identity $$ \sum_{i=0}^{[n / 2]}\left(\begin{array}{l} n \\ 2 i \end{array}\right) p^{2 i} q^{n-2 i}=\frac{1}{2}\left[(p+q)^{n-}+(q-p)^{n}\right] $$ where \([n / 2]\) is the largest integer less than or equal to \(n / 2\). Compare this exercise with Theoretical Exercise 15 of Chapter \(3 .\)

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A newsboy purchases papers at 10 cents and sells them at 15 cents. However, he is not allowed to return unsold papers. If his daily demand is a binomial random variable with \(n=10, p=\frac{1}{3}\), approximately how many papers should he purchase so as to maximize his expected profit?

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